Find the equation of the plane that contains the point $(1, - 1, 2)$ and is perpendicular to each of the planes $2 x + 3 y - 2 z = 5$ and $x + 2 y - 3 z = 8$

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Solution

we have,

drs of required plane $= \to n = n_{1} \times n_{2}$

$n_{1} \to 2 x + 3 y - 2 z = 5 \to (2, 3, - 2)$

$n_{2} \to x + 2 y - 3 z = 8 \to (1, 2 - 3)$

$\to n = ∣ ∣ ∣ ∣ \begin{matrix} i & j & k 2 & 3 & - 2 1 & 2 & - 3 \end{matrix} ∣ ∣ ∣ ∣$

$\Rightarrow n = (- 5, 4, 1)$

Equation of the required plane is $- 5 x + 4 y + z + d = 0$

Given that plane contains the point $(1, - 1, 2)$

it should satisfy the equation of the plane.

$- 5 (1) + 4 (- 1) + 2 + d = 0$

$- 5 - 4 + 2 + d = 0$

$d = 7$

Therefore the required equation is,
$∴ - 5 x + 4 y + z + 7 = 0$

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Similar questions

(1, - 1, 2)

x + 2 y + 2 z = 5

3 x + 3 y + 2 z = 8

(- 1, 3, 2)

x + 2 y + 2 z = 5, 3 x + 3 y + 2 z = 8

(2, 4. - 1)

3 x - 4 y + 2 z - 7 = 0

x + 2 y - 3 z + 2 = 0.

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